We may represent a general electromagnetic plane wave by (real part of the complex exponentials): $$\vec E = \vec E_0\exp(i\vec k \cdot \vec r - i \omega t) \quad\text{&}\quad\vec B = \vec B_0\exp(i\vec k \cdot \vec r - i \omega t)$$ Show that Faraday's law $$\partial_t \vec B=-\nabla \times \vec E\tag{1}$$ becomes $$\omega \vec B_0=\vec k \times \vec E_0\tag{2}$$
I know I can write $$\partial_t \vec B=-i \omega\exp(i\vec k \cdot \vec r - i \omega t)\vec B_0$$ for the LHS of $(1)$
Letting $f=\exp(i\vec k \cdot \vec r - i \omega t)$ then $\partial_t \vec B=-i \omega f \vec B_0$ and $\vec E = f \vec E_0$ and the RHS of $(1)$ is $$\begin{align} -\nabla \times \vec E &=-\left(\nabla\times(f \vec E_0)\right)\\&=-\left(\nabla f \times \vec E_0 + f (\nabla \times \vec E_0)\right)\\&=-\left(i f\vec k \times \vec E_0+f(\nabla \times \vec E_0)\right)\end{align}$$
The only way I can reach $(2)$ is iff $$\nabla \times \vec E_0=\vec 0$$ which would be the case under the assumption that $\vec E_0$ and $\vec B_0$ are constant vectors.
The question did not stipulate this and in the answer there is no mention of it:
Is my assumption correct?